a bag contains title with the letters a-r-i-t-h-m-e-t-i-c. Amelia chooses a title without looking and doesn't replace it. She chose a secnd title without looking. what is the probability that she will choose the letter I both times?
* 1/25
* 1/45
* 2/45
* 2/55
The probability of choosing the letter "I" on the first draw is 2/10 since there are two "I"s out of a total of 10 letters. Since Amelia does not replace the chosen title before drawing again, the probability of choosing "I" on the second draw is 1/9 since there is now only one "I" left out of a total of 9 remaining letters. So the probability of choosing "I" twice in a row is (2/10) x (1/9) = 2/45. Therefore, the answer is (C) 2/45.
There are 40 entries i the sciece fair. How many ways ca the etries fiish in first,second, and third place?
240
1,480
12,640
59,280
There are 40 options for first place. Once first place is decided, there are 39 options for second place. Once first and second place are decided, there are 38 options for third place. Therefore, the total number of ways the entries can finish in first, second, and third place is:
40 x 39 x 38 = 59,280
So the answer is (D) 59,280.
thank you
You're welcome! If you have any more questions, feel free to ask.
When buying a new dress, you have a choice of 3 different lengths,5 different colors, and 2 different styles. How many choices are there for 1 dress?
13
15
30
32
To find the total number of choices for one dress, we need to multiply the number of options for length, color, and style:
3 x 5 x 2 = 30
Therefore, there are 30 different choices for one dress, so the answer is (C) 30.
How many different ways can a band teacher select the lead and co-lead trumpet player from a group of 12 trumpet players?
132
264
312
448
There are 12 options for the first trumpet player (the lead player) and 11 options remaining for the second trumpet player (the co-lead). However, since order does not matter between the lead and co-lead players, we need to divide by 2 to avoid overcounting. Therefore, the total number of ways the band teacher can select the lead and co-lead trumpet player is:
(12 x 11)/2 = 132
So the answer is (A) 132.
Below are the results of tossing a number cube 8 times. Find the experimental probability of tossing an even number.
3,4,1,2,5,1,6
1/4
3/8
1/2
5/8
Out of the 8 tosses, the even numbers that were rolled were 4, 2, and 6. Therefore, the number of times an even number was rolled is 3. Since each toss has an equal probability of being an even number or an odd number, the total number of possible outcomes is 8. So the experimental probability of rolling an even number is:
3/8
Therefore, the answer is (B) 3/8.
Write the number of permutations in the factorial form. Then simplify. how many differnt ways can you and four of your friends sit in the back seat of a limousine?
4!; 24
4!; 120
5!; 120
5!; 720
The number of permutations (ways to seat 5 people) can be written as:
5!
which is read as "5 factorial." To simplify, we can calculate the value of 5!:
5! = 5 x 4 x 3 x 2 x 1 = 120
Therefore, there are 120 different ways that you and four of your friends can sit in the back seat of a limousine. Thus, the answer is (B) 4!; 120.
You want to draft a four-player tennis team. There are eight players to choose from. How many different teams can you form?
70
64
48
336
The number of ways to form a team of 4 players from a group of 8 is given by the combination:
8 C 4 = (8!)/(4!(8-4)!) = (8!)/(4!4!) = (8x7x6x5)/(4x3x2x1) = 70
Therefore, there are 70 different teams that can be formed. So the answer is (A) 70.
How many different arrangements can be made with the letters from the word MATH?
12
24
106
1,210
There are 4 letters in the word MATH, so the number of different arrangements we can make is given by:
4! = 4 x 3 x 2 x 1 = 24
Therefore, there are 24 different arrangements that can be made with the letters from the word MATH. So the answer is (B) 24.
Eric has two identical number cubes. The faces of each cube are numbered 1-6. Eric will roll both of the number cubes at the same time. What probability that both of the numbers showing face-up will be multiples of 3?
2/9
1/3
1/6
1/9
Not one of the chocies
My apologies. Let me calculate the correct answer for you.
Out of these 36 possible outcomes, there are 4 outcomes where both cubes show a multiple of 3: (3, 3), (3, 6), (6, 3), and (6, 6).
Therefore, the probability of both cubes showing a multiple of 3 is:
4/36 = 1/9
Therefore, the answer is (D) 1/9.
There are two multiples of 3 on each cube: 3 and 6. There are 6 possible outcomes for rolling the first cube and 6 possible outcomes for rolling the second cube, so the total number of possible outcomes is 6 x 6 = 36.
Out of these 36 possible outcomes, there are 2 outcomes where both cubes show a multiple of 3: (3, 3) and (6, 6).
Therefore, the probability of both cubes showing a multiple of 3 is:
2/36 = 1/18
This can be simplified to:
1/3 x 1/6 = 1/18
Therefore, the answer is (D) 1/18.